Question

DEFG is quadrilateral such that DF divides it into two parts of equal areas. prove that diagonal DF bisect GE.

Answer 1

Area of DEFG = Area of $\Delta DEF$ + Area of $\Delta FGD$
Area of DEFG = Area of $\Delta DEF$ + Area of $\Delta DEF$ ($\because$ diagonal divides quadrilateral into 2 parts of equal area)
$\therefore \Delta DEF\cong \Delta FGD$
DE = GF (corresponding sides of congruent triangles)
$\angle DFE=\angle FDG$ (corresponding angles of a congruent triangle)
But, these angles are in a position of alternate interior angles
$\therefore$ DE || GF
Similarly, $\angle EDF=\angle GFD$ and DG ||EF
So DEFG is a parallelogram
$\because \angle EDO=\angle GFO$
DE = FG
$\angle DEO=\angle FGO$
$\therefore \Delta DOE\cong \Delta FOG$ ( by A-S-A property)
DO = FO (corresponding sides of congruent triangles)
OE = OG (corresponding sides of congruent triangles)
Its proved the two diagonals bisect each other.